Why are the gamma-matrices invariant?

[PatrickUrania]

Hi,
I've been studying Dirac's theory of fermions. A classic topic therein is the proof that the equation is covariant. Invariably authors state that the gamma-matrices have to be considered constants: they do not change under a Lorentz-transformation. I am looking for the reason behind this. It seems to me that if you consider them a vector and the wave-function a scalar then all works out OK. The scalar, vector, tensor, pseudoscalar and pseudovector constructed from the gamma-matrices, the wave-function and its adjoint all have the same value as in the classical approach and transform in the correct fashion. What am I missing?

 

[dextercioby]

Have you asked yourself the question: what does a Lorentz transformation act on?

 

[stevendaryl]

An interesting paper discussing this was written (I think) by Physics Forums' own Demystifier:
http://arxiv.org/pdf/1309.7070.pdf

In this paper, the author claims that the physical content of the Dirac equation is unchanged, whether you view spinors as transforming under Lorentz transformations and gamma matrices as being constant, or vice-versa.

 

[PatrickUrania]

@Dextercloby: As far as I can see, the Lorentz transformation acts on everything that lives in spacetime, but only on those objects. That implies it obviously acts on the momentum operator. By the same token it clearly should have an action on the gamma-matrices as they, at least partially, live in spacetime as well. Since the wavefunction has a spacetime point as its argument, that event has to be transformed. The action on the wavefunction itself is much less clear. Looking at it as a set of complex scalars does not seem unreasonable. In that view, the wavefunction exists in a ℂ⁴ space that is separate from spacetime. The gamma-matrices have 2 indices in that space as well.

 

[PatrickUrania]

@stevendaryl: I had read the article but was not aware that the author had a link to this forum. It indeed exactly describes what I'm asking here. What bothers me is that, given that there is a much simpler way to use spinors, why would anyone use the traditional way? I have been looking at some literature concerning tetrads (or vierbeins), that indeed seems to go in that direction. There is also the Dirac-Fock-Weyl equation that takes the Dirac equation to curved space and that - as far as I can see - again seems to point in that direction. Caveat: I am by no means an expert in quantum mechanics on curved spaces. However, the very vast majority still use the traditional way. I can only imagine that there is a good reason to do so, but so far I haven't seen it.

 

[Demystifier]

@stevendaryl: I had read the article but was not aware that the author had a link to this forum. It indeed exactly describes what I'm asking here. What bothers me is that, given that there is a much simpler way to use spinors, why would anyone use the traditional way? I have been looking at some literature concerning tetrads (or vierbeins), that indeed seems to go in that direction. There is also the Dirac-Fock-Weyl equation that takes the Dirac equation to curved space and that - as far as I can see - again seems to point in that direction. Caveat: I am by no means an expert in quantum mechanics on curved spaces. However, the very vast majority still use the traditional way. I can only imagine that there is a good reason to do so, but so far I haven't seen it.

One obvious reason why the simple formalism (I developed in the recent paper you have seen) is not widely used, is that this formalism is not widely known. (The related formalism with tetrads is better known, but is not so simple.)

Another reason is that in most practical calculations it doesn't really matter which formalism you use, because in practice one rarely needs to transform a quantity from one Lorentz frame to another. Instead, one simply works in the fixed laboratory frame, in which there is no difference between the two approaches.

My formalism is much simpler when you want to prove Lorentz covariance of the Dirac equation. But once you prove it (in either a simple or a complicated way), you don't longer need to worry much about it.

In fact, the true reason why I developed this formalism in the first place is because I needed some new bilinear covariant combinations mentioned in the paper, which cannot easily be constructed in the standard formalism. But these new bilinear combinations appear in a certain non-standard formulation of QM, not in standard QM/QFT.

 

[stevendaryl]

In fact, the true reason why I developed this formalism in the first place is because I needed some new bilinear covariant combinations mentioned in the paper, which cannot easily be constructed in the standard formalism. But these new bilinear combinations appear in a certain non-standard formulation of QM, not in standard QM/QFT.

Just curious: is that Hestene's spacetime algebra approach?

 

[Demystifier]

Just curious: is that Hestene's spacetime algebra approach?

It is not, but maybe there is a relation to it. I would need to refresh my knowledge about the Hestenes approach and think about it.

 

[vanhees71]

In the usual operator formulation of relativistic QFT the fields are operators, and a Lorentz transformation is realized as a unitary (ray) representation on that fields. The $\gamma$ matrices are complex-number matrices and thus the QFT operator doesn't do anything to them.

 

[PatrickUrania]

In the usual operator formulation of relativistic QFT the fields are operators, and a Lorentz transformation is realized as a unitary (ray) representation on that fields. The $\gamma$ matrices are complex-number matrices and thus the QFT operator doesn't do anything to them.

Your respons is indeed the standard one. However, why exactly do the γ-matrices have to be just complex numbers? As was explained above, considering them as a vector and the wavefunction as a scalar leads to the same physics, only without the need to introduce another representation of the Lorentz transformation. The article mentioned above explains this in detail. Is there a compelling reason why the γ-matrices have to be just complex numbers?

 

[Demystifier]

Your respons is indeed the standard one. However, why exactly do the γ-matrices have to be just complex numbers? As was explained above, considering them as a vector and the wavefunction as a scalar leads to the same physics, only without the need to introduce another representation of the Lorentz transformation. The article mentioned above explains this in detail. Is there a compelling reason why the γ-matrices have to be just complex numbers?

In curved spacetime, it's not only that $\gamma^{\mu}$ is a vector, but is also a function $\gamma^{\mu}(x)$ depending on the spacetime point $x$. So no, the γ-matrices do not have to be just complex numbers.

In fact, the γ-matrices satisfy the algebra
$$\{\gamma^{\mu}(x), \gamma^{\nu}(x) \} =2g^{\mu\nu}(x)$$
where $g^{\mu\nu}(x)$ is the spacetime metric. In Minkowski spacetime this can be reduced to
$$\{\gamma^{\mu}, \gamma^{\nu} \} =2\eta^{\mu\nu}$$
Hence γ-matrices can be chosen to be just numbers only when the metric tensor is just numbers.

 

[vanhees71]

The $\gamma$ matrices cannot be expressed in terms of field operators and are thus c-number valued matrices.

In GR spinors are, as any fields, local objects, i.e., they have to be defined with respect to comoving vierbeins.

 

[Demystifier]

The $\gamma$ matrices cannot be expressed in terms of field operators and are thus c-number valued matrices.

In GR spinors are, as any fields, local objects, i.e., they have to be defined with respect to comoving vierbeins.

In quantum gravity the vierbein may also be quantized i.e. become a field operator. In this case $\gamma^{\mu}(x)$ also becomes a field operator.

 

[Demystifier]

The above gave me a wild idea. Consider a quantity (operator, matrix, or something like that) $I(x)$ satisfying
$$I^2(x)=g(x)$$
where $g(x)$ is the determinant of the metric tensor with signature (-+++). In Minkowski spacetime this reduces to
$$I^2(x)=-1$$
so $I$ can be represented by the imaginary unit $I=i$. But in curved spacetime $I(x)\neq i$.

What if in quantum theory we replace $i$ with $I(x)$? Could it have something to do with merging quantum theory with gravity?

As I said, at this level this is only a wild speculation. But has anybody thought about something like that?

 

[vanhees71]

That's a funny idea. I've no clue, if anybody has thought about such a thing, but I'm no expert in attempts to create a consistent QT of gravity.

 

[stevendaryl]

I'm not sure whether it is a different way of thinking of the same thing, or not, but in the Hestenes "geometric algebra" approach to the Pauli equation or the Dirac equation, the gamma matrices (or the Pauli matrices) are not treated as matrices, at all, but are just considered vectors. That is, Hestenes considers $\gamma^\mu$ to be a set of 4 basis vectors, rather than 4 components of a single vector.

 

[vanhees71]

But $\gamma^{\mu}$ cannot be just a number, because it fulfills the Clifford algebra $[\gamma^{\mu},\gamma^{\nu}]_+=2 g^{\mu \nu}$. Of course you can think of it in terms of abstract objects as well as you can think of the Dirac spinors being just abstract objects. Then the usual physicists' objects are components of spinors and matrix elements of Clifford-algebra elements defined as linear maps on the spinors.

 

[PatrickUrania]

In curved spacetime, it's not only that $\gamma^{\mu}$ is a vector, but is also a function $\gamma^{\mu}(x)$ depending on the spacetime point $x$. So no, the γ-matrices do not have to be just complex numbers.

In fact, the γ-matrices satisfy the algebra
$$\{\gamma^{\mu}(x), \gamma^{\nu}(x) \} =2g^{\mu\nu}(x)$$
where $g^{\mu\nu}(x)$ is the spacetime metric. In Minkowski spacetime this can be reduced to
$$\{\gamma^{\mu}, \gamma^{\nu} \} =2\eta^{\mu\nu}$$
Hence γ-matrices can be chosen to be just numbers only when the metric tensor is just numbers.

I found this article by A. Weldon: "Fermions without vierbeins in curved space-time" (cds.cern.ch/record/466101/files/0009086.pdf or on arxiv)> There is also some explanation of these ideas on: http://www.qgf.uni-jena.de/gk_quantenmedia/downloads/Monitoring2012/talk_lippoldt.pdf. This looks like the correct generalisation of these ideas in curved space-time.
From all the above I think that considering the γ-matrices as a true spacetime vector is not only justified but also greatly simplifies the interpretation of what a spinor actually is.

 

[samalkhaiat]

Hi,
I've been studying Dirac's theory of fermions. A classic topic therein is the proof that the equation is covariant. Invariably authors state that the gamma-matrices have to be considered constants: they do not change under a Lorentz-transformation. I am looking for the reason behind this.

See the theorem below.

It seems to me that if you consider them a vector and the wave-function a scalar then all works out OK. The scalar, vector, tensor, pseudoscalar and pseudovector constructed from the gamma-matrices, the wave-function and its adjoint all have the same value as in the classical approach and transform in the correct fashion. What am I missing?

You are missing a lot. Let me list few for you.
1) The Lorentz group is a matrix Lie group. As such, $SO(1,3)$ sees every index carried by a given object and transforms that index in the appropriate representation matrix. So, for $\Gamma^{\mu}_{ab}$ the index $a$ transforms in the spinor representation space $S_{4}$, $b$ in the conjugate spinor representation $\bar{S}_{4}$ and $\mu$ in the vector representation $D^{(1/2,1/2)}$. This means that $$\Gamma^{\mu}_{ab} \in S_{4}(\Sigma) \oplus \bar{S}_{4}(\bar{\Sigma}) \oplus D^{v}(M) .$$
So, the action of Lorentz algebra (infinitesimal transformation) on $\Gamma^{\mu}$ is, therefore, given by
$$\begin{align*} \delta^{\mu\nu} \Gamma^{\rho}_{ac} &= \Sigma^{\mu\nu}_{ab}\Gamma^{\rho}_{bc} - \bar{\Sigma}^{\mu\nu}_{cb}\Gamma^{\rho}_{ab} + (M^{\mu\nu})^{\rho}_{\sigma}\Gamma^{\sigma}_{ac} \\ &= \Sigma^{\mu\nu}_{ab}\Gamma^{\rho}_{bc} - \Gamma^{\rho}_{ab}\Sigma^{\mu\nu}_{bc} + (M^{\mu\nu})^{\rho}_{\sigma}\Gamma^{\sigma}_{ac} \\ &= [ \Sigma^{\mu\nu} , \Gamma^{\rho} ]_{ac} + (M^{\mu\nu})^{\rho}_{\sigma}\Gamma^{\sigma}_{ac} . \end{align*}$$
Now we can prove the theorem:
The matrices $\Sigma^{\mu\nu} = - \frac{i}{4}[\Gamma^{\mu} , \Gamma^{\nu} ]$ generate the spinor representation of $SO(1,3)$ if and only if $\delta \Gamma = 0$.
Proof:
i) Using the Dirac algebra $\big \{ \Gamma^{\mu} , \Gamma^{\nu} \big \} = 2 \eta^{\mu\nu}$, we can calculate the commutator
$$\begin{align*} [\Sigma^{\mu\nu} , \Gamma^{\rho} ]_{ac} &= i \left( \eta^{\mu\rho} \Gamma^{\nu}_{ac} - \eta^{\nu\rho} \Gamma^{\mu}_{ac} \right) \\ &= i \left( \eta^{\mu\rho}\delta^{\nu}_{\sigma} - \eta^{\nu\rho}\delta^{\mu}_{\sigma} \right) \Gamma^{\sigma}_{ac} \\ &= - (M^{\mu\nu})^{\rho}_{\sigma} \Gamma^{\sigma}_{ac} . \end{align*}$$
In the last equality we introduced the generator matrices $M^{\mu\nu}$ of the vector representation
$$(M^{\mu\nu})^{\rho}_{\sigma} = - i \left( \eta^{\mu\rho}\delta^{\nu}_{\sigma} - \eta^{\nu\rho}\delta^{\mu}_{\sigma} \right) .$$
Thus
$$\delta^{\mu\nu}\Gamma^{\rho}_{ac} = - (M^{\mu\nu})^{\rho}_{\sigma} \Gamma^{\sigma}_{ac} + (M^{\mu\nu})^{\rho}_{\sigma} \Gamma^{\sigma}_{ac} = 0 .$$
ii) Conversely, starting from $\delta \Gamma = 0$, we arrive at the generator matrices of the spinor representation $\Sigma^{\mu\nu} = (-i/4) [ \Gamma^{\mu} , \Gamma^{\nu}]$ as the Clifford solution of
$$[\Sigma^{\mu\nu}, \Gamma^{\rho}]_{ac} = i \left( \eta^{\mu\rho} \Gamma^{\nu}_{ac} - \eta^{\nu\rho} \Gamma^{\mu}_{ac} \right) .$$
This is nothing but the infinitesimal version of
$$S^{-1}(\Lambda) \Gamma^{\mu}S(\Lambda) = \Lambda^{\mu}{}_{\nu}\Gamma^{\nu} , \ \ \ \ \ \ (1)$$
which follows from the form-invariance of the Dirac equation. Manny textbooks say that (1) is the Lorentz transformation law for Dirac’s gammas. This is wrong, because (as we have just shown) (1) follows from the invariance of the $\Gamma$’s under Lorentz transformations. One should mention that the above theorem can be proved for any matrix Lie group.
2) Quantum theory can be made independent of a frame if the Hilbert space of the theory carries a unitary representation of the Poincare’ group, i.e.,
$$|\psi_{\alpha} \rangle \to U(\Lambda , a) |\psi_{\alpha}\rangle , \ \ \forall |\psi_{\alpha}\rangle \in \mathcal{H} ,$$ or, in terms of multi-component wave-function $$\psi^{'}_{\alpha}(x) = \langle x |U(\Lambda , a)|\psi \rangle = \mathcal{U}(\Lambda , a) \psi_{\alpha}(x) . \ \ \ \ (2)$$ But the Hilbert space of relativistic n-component wave functions is simply the space $$\mathcal{H} \cong \mathcal{F}(\mathbb{R}^{4}) \otimes \mathbb{C}^{n} ,$$ where $\mathcal{F}(\mathbb{R}^{4})$ is the space of functions on Minkowski spacetime. So, the Poincare’ group transforms the argument of the Dirac wavefunction (which comes from $\mathcal{F}(\mathbb{R}^{4})$) and mixes its components (which come from $\mathbb{C}^{4}$) :
$$\psi^{'}_{\alpha} (x) = S_{\alpha \beta}(\Lambda) \psi_{\beta}(\Lambda^{-1}(x-a)) . \ \ \ \ \ (3)$$ So, saying that the Dirac wavefunction transforms like scalar function will simply mean the followings very bad things:
(i) the operator $U(\Lambda , a)$ cannot be a unitary operator within the time independent scalar product of the Hilbert space of solutions of the Dirac equation, and (ii) the infinitesimal generator of $\mathcal{U}(\Lambda, a)$ does not contain the spin-1/2 matrices.

 

[PatrickUrania]

Thanks for the response. This is the first time I see an actual reason for this behavior. I don't understand everything though. Could you point me to a reference that explains this in some more detail?

 

[vanhees71]

You find an excellent treatment about the unitary representations of the Poincare group in

S. Weinberg, Quantum Theory of Fields, vol. I, Cambridge University Press (1995)

It's also in my QFT manuscript

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

 

[PatrickUrania]

You find an excellent treatment about the unitary representations of the Poincare group in

S. Weinberg, Quantum Theory of Fields, vol. I, Cambridge University Press (1995)

It's also in my QFT manuscript

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Thanks for the information.
I have been looking at your manuscript and I noted that in appendix B you show that the Lorentz group O(1,3) has 4 covering groups, each with two spinor representations, but in passing you note that there is also the trivial representation. That looks a lot like what I was thinking about: don't transform the spinors. Why would that not be a valid choice?
Also I've been looking at the proof that the Dirac equation leads to a unitary wavefunction. The original proof does not seem to depend upon the transformation properties of the wavefunction or the γ-matrices. The resulting equation:
∂_μ([BAR]Ψ[/BAR] γ^μ ψ) = 0
is valid in all frames in both formalisms (it depends on the transformation of a bilinear form and all those transform in exactly the same way whatever formalism you choose). So I don't see an evident reason why unitarity would be lost. Is there a simple way to show where this goes wrong?

 

[samalkhaiat]

Thanks for the response. This is the first time I see an actual reason for this behavior. I don't understand everything though. Could you point me to a reference that explains this in some more detail?

Any decent textbook on the representation theory of the Clifford group $\mbox{Pin}(m,n)$ and/or $\mbox{Spin}(m,n)$ should say something about the behaviour of the $\Gamma$’s. I believe H. Georgi’s text “Lie Algebras in Particle Physics” has detailed account of the spin groups and their representations.
You can also ask me about the things you did not understand.

 

[samalkhaiat]

.. you note that there is also the trivial representation. That looks a lot like what I was thinking about: don't transform the spinors. Why would that not be a valid choice?
Also I've been looking at the proof that the Dirac equation leads to a unitary wavefunction. The original proof does not seem to depend upon the transformation properties of the wavefunction or the γ-matrices.
So I don't see an evident reason why unitarity would be lost. Is there a simple way to show where this goes wrong?

A key requirement on the scalar product of the Hilbert space of solutions (of the Dirac equation) is that it be conserved under the time evolution. This is closely related to the unitarity requirement that probabilities sum to one. The appropriate scalar product for Dirac theory is given by
$$(\Phi , \Psi ) = \int_{\sigma} d \sigma^{\mu}(x) \ \overline{\Phi}(x) \Gamma_{\mu} \Psi (x) ,$$ where $\sigma (x)$ is an arbitrary space-like hypersurface. Indeed, this scalar product is conserved in time if $\Phi$ and $\Psi$ solve the Dirac equation:
$$\begin{align*} \frac{\delta}{\delta \sigma (x)} (\Phi , \Psi ) &= \partial^{\mu} \left( \overline{\Phi}(x) \Gamma_{\mu} \Psi (x) \right) \\ &= \left( \partial^{\mu}\overline{\Phi}(x) \Gamma_{\mu}\right) \Psi(x) + \overline{\Phi}(x) \left( \Gamma_{\mu}\partial^{\mu} \Psi (x) \right) \\ &= 0 . \end{align*}$$
Thus, the scalar product is independent of $\sigma$, hence we may evaluate it on the plane $t = \mbox{const.}$ Having defined the appropriate scalar product, we will now show that the mapping $|\Psi \rangle \to U(\Lambda , a)|\Psi \rangle$ is unitary if the following non-trivial transformations hold
$$\Psi^{'}_{\alpha}(x) = \mathcal{U}(\Lambda , a)\Psi_{\alpha}(x) = S_{\alpha\beta}(\Lambda) \Psi_{\beta}\left( (\Lambda , a )^{-1}x \right) ,$$
with $S(\Lambda) \in SO(1,3)$ and $(\Lambda , a)^{-1} \in ISO(1,3)$ are given by
$$S^{-1}(\Lambda) \Gamma_{\mu}S(\Lambda) = \Lambda^{\nu}{}_{\mu}\Gamma_{\nu} ,$$ $$(\Lambda , a )^{-1}x = (\Lambda^{-1} , - \Lambda^{-1}a ) x = \Lambda^{-1}(x-a).$$ The words “non-trivial”, “non-singlet”, “non-scalar”, and “$S(\Lambda) \neq I$” all mean the same thing, that is “the representation space is not $D^{(0,0)}$”.
As I said in the other post, the action of $\mathcal{U}(\Lambda , a) \in ISO(1,3)$ on the multi-component wavefunction $\Psi_{\alpha}(x) \in \mathcal{F}(\mathbb{R}^{4}) \otimes \mathbb{C}^{4}$ consists of: the element $(\Lambda , a)^{-1} = (\Lambda^{-1}, -\Lambda^{-1}a) \in ISO(1,3)$ acting on the spacetime argument of the wavefunction, and the matrix $S(\Lambda) \in SO(1,3)$ mixing its components.
Okay, let us now show that the above transformations mean that $U(\Lambda , a)$ is a unitary operator.
$$\begin{align*} \left( U \Phi , U \Psi \right) &= \int_{\sigma} d\sigma^{\mu}(x) \ \overline{\left(\mathcal{U}\Phi\right)}(x) \ \Gamma_{\mu} \ \left( \mathcal{U}\Psi \right) (x) \\ &= \int_{\sigma} d\sigma^{\mu}(x) \ \overline{\Phi}\left((\Lambda , a)^{-1}x \right) \ S^{-1}(\Lambda) \Gamma_{\mu} S(\Lambda) \ \Psi \left( (\Lambda , a)^{-1}x \right) \\ &= \int_{\sigma} d\sigma^{\mu}(x) \ \overline{\Phi}\left((\Lambda , a)^{-1}x \right) \ \Lambda^{\nu}{}_{\mu} \ \Gamma_{\nu} \ \Psi \left( (\Lambda , a)^{-1}x \right) \\ &= \int_{\sigma} d\left(\Lambda^{\nu}{}_{\mu}\sigma^{\mu}(x)\right) \ \overline{\Phi}\left((\Lambda , a)^{-1}x \right) \ \Gamma_{\nu} \ \Psi \left( (\Lambda , a)^{-1}x \right) \\ &= \int_{\sigma} d\sigma^{\nu}\left((\Lambda , a)^{-1}x \right) \ \overline{\Phi}\left((\Lambda , a)^{-1}x \right) \ \Gamma_{\nu} \ \Psi \left( (\Lambda , a)^{-1}x \right) \\ &= \int_{\sigma} d\sigma^{\mu}(y) \ \overline{\Phi}(y) \Gamma_{\mu} \Psi(y) \\ &= \left(\Phi , \Psi \right) . \end{align*}$$
Thus, $U^{\dagger}U = UU^{\dagger} = 1, \ \mbox{qed}$.
Next, we show that the infinitesimal generator of $\mathcal{U}$ is the sum of orbital angular momentum and spin angular momentum. This follows by considering the infinitesimal version of the non-trivial transformation of Dirac wavefunction
$$\begin{align*} (1 - i \epsilon J) \Psi_{\alpha}(x) &= (I_{4} + \epsilon \Sigma)_{\alpha\beta}\Psi_{\beta}(x - \epsilon \omega x) \\ &= (\delta_{\alpha\beta} + \epsilon \Sigma_{\alpha\beta}) \left( \Psi_{\beta}(x) - \epsilon (\omega x)^{\mu}\partial_{\mu}\Psi_{\beta}(x) \right) . \end{align*}$$
Thus
$$J = - i (\omega x)^{\mu}\partial_{\mu} + i \Sigma .$$ This is nothing but the relativistic generalization of the total angular momentum $J = L + S$ in QM.
So, I was not telling you a lie when I said “very bad things” happen if the Dirac wavefunction transforms in the trivial (scalar, singlet) representation of the Lorentz group: Neither $U(\Lambda , a)$ is unitary, nor the spin matrix $\Sigma$ is non-zero.

 

[dextercioby]

Hi Sam,

technically, $S(\Lambda)$ is always in the finite dimensional spinor space, it's not an element of the (restricted) Lorentz group. Likewise, $\mathcal{U}(\Lambda, a)$ is in the (rigged) Hilbert space of the representation, thus not in ISO(1,3).

 

[PatrickUrania]

A key requirement on the scalar product of the Hilbert space of solutions (of the Dirac equation) is that it be conserved under the time evolution. This is closely related to the unitarity requirement that probabilities sum to one.

This part also works if you consider the γ-matrices as a vector and the wavefunction a scalar.
However:

Next, we show that the infinitesimal generator of $\mathcal{U}$ is the sum of orbital angular momentum and spin angular momentum. This follows by considering the infinitesimal version of the non-trivial transformation of Dirac wavefunction

This does indeed seem to require a non-trivial transformation of the wavefunction. I'll have to study on that some more.
It's a pity that the vast majority of references just seem to gloss over this part. On the other hand, it could just as well be me missing a lot of background.
Thanks again for your patience and helping me to fill in some blanks.

 

[samalkhaiat]

Hi Sam,

technically, $S(\Lambda)$ is always in the finite dimensional spinor space, it's not an element of the (restricted) Lorentz group. Likewise, $\mathcal{U}(\Lambda, a)$ is in the (rigged) Hilbert space of the representation, thus not in ISO(1,3).

Hi,
Yes you are right. However, by $S(\Lambda) \in SO(1,3)$, I meant to say that $S(\Lambda)$ is a finite-dimensional matrix representation of $SO(1,3)$, i.e., the map $\pi : \Lambda \to S(\Lambda)$ is a $SO(1,3)$ group homomorphism:
$$S(\Lambda_{1}) S(\Lambda_{2}) = S(\Lambda_{1}\Lambda_{2}) ,$$
with the “spinor space” $\mathcal{S}^{4}$ being the representation space. So the pair $(\pi , \mathcal{S}^{4})$ is a representation of $SO(1,3)$.
Similarly,
$$\mathcal{U}(\Lambda_{1},a_{1}) \mathcal{U}(\Lambda_{2},a_{2}) = \mathcal{U}(\Lambda_{1} \Lambda_{2}, \Lambda_{1}a_{2} + a_{1}) .$$
So, the operator $\mathcal{U}$ forms an infinite-dimensional unitary representation of $ISO(1,3)$ with induced action on wavefunctions in $\mathcal{H}$.

 

[samalkhaiat]

This part also works if you consider the γ-matrices as a vector and the wavefunction a scalar.

No, it does not. You have to invent a new time-independent scalar product which, for a scalar wavefunctions, does not involve the gamma matrices. Look up the scalar product for the complex KG wavefunction.

 

[Demystifier]

No, it does not. You have to invent a new time-independent scalar product which, for a scalar wavefunctions, does not involve the gamma matrices. Look up the scalar product for the complex KG wavefunction.

When wave function is treated as a scalar and gamma matrices as a vector, one can introduce two time-independent scalar products. One is the Dirac scalar product and the other is the Klein-Gordon scalar product. Both are of the form
$$\int_{\Sigma} d\Sigma^{\mu} j_{\mu}$$
where $j_{\mu}$ is either the Dirac of Klein-Gordon current, both described in http://lanl.arxiv.org/abs/1309.7070 (Eqs. (48) and (50))

 

[samalkhaiat]

When wave function is treated as a scalar and gamma matrices as a vector,

This is becoming a kid’s stuff. I proved no such freedom is allowed. The Lorentz group (and all other spin groups) sees $\Gamma^{\mu}$ as an invariant object, and treats Dirac’s wavefunction as non-scalar.

one can introduce two time-independent scalar products.

Two scalar products on one Hilbert space?

One is the Dirac scalar product and the other is the Klein-Gordon scalar product.

Which one is yours? The Hilbert space of Dirac theory is different from the Hilbert space of the K-G theory.

Both are of the form
$$\int_{\Sigma} d\Sigma^{\mu} j_{\mu}$$
where $j_{\mu}$ is either the Dirac of Klein-Gordon current,

This is the Lorentz invariant charge, i.e., probability or the square of the norm $(\phi , \phi)$. Scalar product is defined for two state vectors $(\phi_{1},\phi_{2})$:
1) The Hilbert space of the K-G theory is equipped with the following time-independent scalar product
$$(\phi_{1},\phi_{2}) = i \int_{\sigma} d\sigma^{\mu}(x) \ \phi^{*}_{1}(x) \overleftrightarrow{\partial_{\mu}} \phi_{2}(x) .$$
With this scalar product, the Poincare’ group acts unitarily only if the K-G wave function transforms as scalar. Indeed, we can easily prove that $(U(\Lambda , a)\phi_{1} , U(\Lambda , a)\phi_{2}) = (\phi_{1},\phi_{2})$ only if
$$\phi^{'}_{i}(x) = \mathcal{U}(\Lambda , a) \phi_{i}(x) = \phi_{i}(\Lambda^{-1}(x-a)) , \ \ \ \ i = 1,2 .$$
2) The Hilbert space of Dirac theory comes with following the scalar product
$$(\chi , \psi ) = \int_{\sigma} d\sigma^{\mu}(x) \ \bar{\chi}(x) \gamma_{\mu} \psi (x) .$$
In a previous post and with this scalar product, I proved that the Poincare group acts by unitary operator only if the Dirac wavefunction transforms in the non-trivial (spinor) representation, i.e., $(U(\Lambda,a)\chi , U(\Lambda ,a)\psi) = (\chi ,\psi)$ only if $S(\Lambda) \neq I$.

both described in http://lanl.arxiv.org/abs/1309.7070 (Eqs. (48) and (50))

i) If you claim that you have a “formalism”, then you should be able to define your time-independent scalar product and prove to us that the Poincare group acts by unitary operator with in your scalar product.
ii) If you claim that $\gamma^{\mu}_{\alpha\beta}$ is a vector and $\psi_{\beta}(x)$ is a scalar, then you should explain to us ,group theoretically, why the Lorentz group (which is a matrix spin group) sees the vector index $\mu$ and gives a blind eye to the spin indices $(\alpha, \beta)$.
As well as many mathematical misunderstanding, your paper fails to address the above issues. I am sorry, but It is simply wrong.

 

[PatrickUrania]

No, it does not. You have to invent a new time-independent scalar product which, for a scalar wavefunctions, does not involve the gamma matrices. Look up the scalar product for the complex KG wavefunction.

You can also look at it this way: Starting from the Dirac equation:
(i γ^μ∂_μ + m) ψ = 0
Then change to a new reference frame where S is the spin ½ representation of the Lorentz-transform Λ:
(i S γ^μ S^-1 Λ_μ^ν∂_ν + m) Sψ = 0
But, the γ-matrices are not unique. If you have a set γ^μ that is consistent with the Clifford-algebra, then, for any unitary matrix U, Uγ^μU^† is also a valid set. Choosing such a set changes the wavefunction ψ to Uψ. This transformation is completely independent of the reference frame.
I now choose such another set of γ-matrices. The Dirac equation in the new frame now becomes:
(i US γ^μ S^-1 U^†Λ_μ^ν∂_ν + m) USψ = 0
But, since S is unitary, so is S^† = S^-1. I choose U=S^†. The Dirac equation, still in the new frame, now simplifies to:
(i γ^μΛ_μ^ν∂_ν + m) ψ = 0
We now have, in the new frame, the same wavefunction as the one we started from and if I call
γ'^ν=γ^μΛ_μ^ν
the γ-matrices in this new frame, this shows they have been transformed as a vector compared to the original ones.
I this is also what is being used in the papers I mentioned above.

 

[Demystifier]

As well as many mathematical misunderstanding, your paper fails to address the above issues. I am sorry, but It is simply wrong.

We already had this discussion, and I asked you a question that you never answered. I will ask you again. How $\gamma^{\mu}$ transforms under general coordinate transformations in curved spacetime? Until you answer that question, you will not convince me that you understand the things sufficiently well.

 

[Demystifier]

ii) If you claim that $\gamma^{\mu}_{\alpha\beta}$ is a vector and $\psi_{\beta}(x)$ is a scalar, then you should explain to us ,group theoretically, why the Lorentz group (which is a matrix spin group) sees the vector index $\mu$ and gives a blind eye to the spin indices $(\alpha, \beta)$.

To understand this, group theory is not enough. Mathematically, one also needs some stuff from differential geometry, such as fibre bundles. Or in a physical language, here one deals with two spaces. One is the spacetime itself (which sees the vector index $\mu$ ), and another is the tangential internal space (which sees the spinor indices $\alpha, \beta$ ). So if you are doing a Lorentz transformation in spacetime, it does not affect the indices in the internal space. Anyway, before claiming again that I am wrong, make sure that you first learned something about fibre bundles, spinors in curved spacetime, etc from the literature.

Or if you don't want to learn new stuff, here is something you might already be familiar with. Are you familiar with Yang-Mills theories in flat spacetime? (Please say you are, otherwise we have a problem.) So in Yang-Mills theories one has spacetime with Lorentz group SO(1,3), and some internal space with an internal gauge group. The internal group, in principle, can be anything, but for QCD it is SU(3), for weak interactions SU(2) etc. The point is that transformations in the internal space are independent on transformations in spacetime. Now what if the internal gauge group is SO(1,3)? Even though it is the same group as for spacetime symmetry, the internal SO(1,3) transformation is independent from the spacetime SO(1,3) transformation.

 

[vanhees71]

Any decent textbook on the representation theory of the Clifford group $\mbox{Pin}(m,n)$ and/or $\mbox{Spin}(m,n)$ should say something about the behaviour of the $\Gamma$’s. I believe H. Georgi’s text “Lie Algebras in Particle Physics” has detailed account of the spin groups and their representations.
You can also ask me about the things you did not understand.

I also like

Sexl, Roman U., Urbandtke, Helmuth K.: Relativity, Groups, Particles, Springer, 2001

very much. It explains the various representations of the Lorentz group(s) and the Poincare group(s) very well, including spinor representations.

 

[samalkhaiat]

..., since S is unitary, so is S^† = S^-1. I choose U=S^†.

No, no no. $S(\Lambda)$ cannot be unitary. It is a finite-dimensional (matrix) representation of the Lorentz group $SO(1,3)$ which is a non-compact Lie group. The fundamental theorem say: a non-compact group has no finite-dimensional unitary representations. Notice that, I always attached the phrase “infinite-dimensional unitary representation” when I dealt with $U(\Lambda , a)$.